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Reptile [31]
3 years ago
7

Severe or sudden changes in Earth's atmosphere can cause entire species of animals to die. When all of the members of species di

e out, it is called
Chemistry
1 answer:
Oduvanchick [21]3 years ago
7 0

Answer: Extinction

Animals on the verge of extinction are considered endangered species. Some could argue that the phrasing "threatened" is synonymous to "endangered". Some classify extinction based on if the species is completely wiped out in the wild, or if the species is completely gone (both in the wild and in zoos for instance). Extinction events are not solely due to the changes in atmosphere, but also due to lack of food or habitat, among other possible factors.

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The average daily diet of an adult woman is 2.4×103kcal. the molar mass of atp is 507 g/mole. part a if all the energy in her di
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To answer this question, we should give the standard free energy change of ATP to ADP conversion that is equal to 30.5 kJ/mol ATP. In this case, 2400 kcal is equal to 10041. 6 kJ. Hence, the mol of ATP produced is 329.23 mol ATP or equal to 166.93 kg ATP. 
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3 years ago
A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric ac
UkoKoshka [18]

Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

pH = pKa + log(\frac{[KF]}{[HF]})

pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04  

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq)  +  F⁻(aq) ⇄   HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

\eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol

\eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

\eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles  

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.  

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)    (1)

The number of moles of HF after the reaction of KF with HCl is:

\eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles

And the concentration of HF after the reaction of KF with HCl is is:

C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L

Now, from the equilibrium of equation (1) we have:

Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]}

Ka = \frac{x^{2}}{0.531 - x}  (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

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I think it should be A
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For example:

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